Answer to Question #327649 in General Chemistry for Missa

Question

Your lab partner accidentally mixed some sodium chloride with your sample of Epsom salts (MgSO₄･7H₂O). You want to make a standard solution of magnesium ion, and this is the only sample of a magnesium salt you have. To determine the amount of magnesium salt in the mixture, you heat 100.00 g to drive off the water of hydration and find that the anhydrous mixture has a mass of 65.75 g. What is the mass in grams of the original salt mixture that you must add to 1.000 L of water to make a 0.1000 M solution of Mg²⁺ ion?

Answer 1

The amount of Mg²⁺ in the final solution: 1.000 L * 0.1000 M = 0.1000 mol

Mass = 0.1000 * 120.366 = 12.0366 (g) of MgSO₄ – needed

Δ = 100.00 – 65.75 = 34.25 g of water or 34.25 / 18.015 = 1.901 mol

The ratio of MgSO₄ to H₂O is 1 : 7 in the salt. So, moles of MgSO₄ = 1.901 / 7 = 0.2716 mol

Mass = 0.2716 * 120.366 = 32.69 (g) or 32.69%

12.0366 g / 0.3269 = 36.82 g of the original mixture